Two identical thin metal plates has charge q₁and q₂ respectively such that $q_{1} > q_{2}$ . The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is

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$\frac{2 (q_{1} - q_{2})}{C}$

$\frac{(q_{1} - q_{2})}{2 C}$

$\frac{(q_{1} - q_{2})}{C}$

$\frac{(q_{1} + q_{2})}{C}$

answer is C.

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Detailed Solution

Electric field between plates $E = \frac{q_{1} - q_{2}}{2 {Aϵ}_{0}}$
$\begin{array}{l} V = Ed = \frac{q_{1} - q_{2} d}{2 {Aϵ}_{0}} \\ V = \frac{q_{1} - q_{2}}{2 C} \end{array}$

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